Please answer and show work:

What time would it take a plane moving at 500milies/hour to travel 4000 miles?

A) 1.25 hours
B) 5 hours
C) 8 hours
D)500 hours

The current flowing through the 30.0 Ω resistor is 2.0 A, which corresponds to option C.

To find the current flowing through the 30 Ω resistor, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R).

Given that the voltage across the circuit is 120.0 V, we can calculate the current flowing through each resistor.

For the 10.0 Ω resistor, using Ohm's Law: I = V/R = 120.0 V / 10.0 Ω = 12.0 A.

For the 20.0 Ω resistor, again using Ohm's Law: I = V/R = 120.0 V / 20.0 Ω = 6.0 A.

Now, let's find the total resistance in the circuit. Since the three resistors are connected in series, we can add them up: R_total = 10.0 Ω + 20.0 Ω + 30.0 Ω = 60.0 Ω.

Next, we can find the current flowing through the 60.0 Ω equivalent resistance. Again, using Ohm's Law: I = V/R = 120.0 V / 60.0 Ω = 2.0 A.

Finally, to find the current flowing through the 30.0 Ω resistor, we can apply Ohm's Law once more: I = V/R = 2.0 A.

Therefore, the current flowing through the 30.0 Ω resistor is 2.0 A, which corresponds to option C.

For more such questions on current visit;

https://brainly.com/question/1100341

#SPJ11

Answer:

Poor conductor metals contain only atoms.

She on side is heated, the atoms vibrate hence passing on their vibration energy to the neighboring atoms. The process is gradual a d continuous

Explanation:

\({}\)

Answer:

Explanation:

Ask a question that can be answered by making observations.

Answer: (APE X) Determine whether experimental observations can provide evidence to support a conclusion

Explanation:

Answer: The capacitor is fully charged when the voltage of the power supply is equal to that at the capacitor terminals. This is called capacitor charging; and the charging phase is over when current stops flowing through the electrical circuit.

Answer:

4161 J/kg·°C

Explanation:

We can use the principle of conservation of energy to solve this problem, which states that the total heat energy in a closed system is constant. The heat lost by the tea is equal to the heat gained by the milk.

Let's first calculate the heat lost by the tea:

Q(tea) = mcΔT

Q(tea) = (0.27 kg)(3910 J/kg·°C)(90.0°C - 85.0°C)

Q(tea) = 6555 J

where m is the mass of tea, c is the specific heat of tea, and ΔT is the change in temperature.

Next, let's calculate the heat gained by the milk:

Q(milk) = mcΔT

Q(milk) = (0.02 kg)(c)(85.0°C - 6.0°C)

Now we can equate the two expressions:

Q(tea) = Q(milk)

6555 J = (0.02 kg)(c)(79.0°C)

Solving for c, we get:

c = 4161 J/kg·°C

Therefore, the specific heat of milk is approximately 4161 J/kg·°C.

Answer:

The current is   \(I  =  1.1434*10^{-5}}\  A\)

Explanation:

From the question we are told that

   The radius of the kite string is  \(R =  2.02 mm =  0.00202 \ m\)

   The  distance it extended upward is   \(D =  0.823 km = 823 \  m\)

   The thickness of the water layer is \(d = 0.506 mm  =  0.000506 \  m\)

   The resistivity is  \(\rho =  159\ \Omega  \cdot m\)

   The potential  difference is  \(V  =   186 MV =  186 *10^{6} \  V\)

Generally the cross sectional area of the water layer is mathematically represented as

      \(A =  \pi r^2\)

Here  r is mathematically represented as

      \(r =  [(R + d ) - R]\)

=>   \(r =  [(0.00202 +  0.000506 ) - 0.00202]\)

=>  \(r =  0.000506\)

=>     \(A = 3.142 *  [0.000506]^2 \)  

=>     \(A = 8.0447*10^{-7}\ m^2 \)  

Generally the resistance of the water is mathematically represented as

    \(R =  \frac{\rho  * D }{A}\)

=>   \(R =  \frac{159  *823 }{8.0447*10^{-7}}\)

=>   \(R = 1.62662 * 10^{11} \  \Omega  \)

Generally the current is mathematically represented as

      \(I  =  \frac{V}{R}\)

=>    \(I  =  \frac{186 *10^{6} }{1.62662 * 10^{11}}\)

=>    \(I  =  1.1434*10^{-5}}\  A\)

Answer:

energy which a body possesses by virtue of being in motion.

Explanation:

hope this help

pls mark as brainliest

Answer: The lowest outside temperature for which the sleeping bag provides adequate protection is approximately 89.61°C below the hiker's skin temperature of 34°C.

Explanation:

To find the lowest outside temperature for which the sleeping bag provides adequate protection, we need to determine the rate of heat loss through conduction and compare it to the heat loss the hiker can safely tolerate.

The rate of heat loss through conduction can be calculated using the formula:

Q = (k * A * ΔT) / d

Where:

Q is the rate of heat transfer (in Watts)

k is the coefficient of thermal conductivity (0.019 Wm¹¹°C-¹ in this case)

A is the surface area (1.3 m² in this case)

ΔT is the temperature difference (in this case, the difference between the skin temperature and the outside temperature)

d is the thickness of the sleeping bag (30 mm, which needs to be converted to meters by dividing by 1000)

Let's plug in the values:

Q = (0.019 * 1.3 * ΔT) / (30 / 1000)

The hiker can safely lose 85 W of heat, so we can set up the equation:

85 = (0.019 * 1.3 * ΔT) / (30 / 1000)

To solve for ΔT, we can rearrange the equation:

ΔT = (85 * (30 / 1000)) / (0.019 * 1.3)

ΔT ≈ 89.61°C

Answer:

kekemeeimdeiddnekem

Explanation:

mdjdjdiddmjd jjeneeiej

Creating an effective system to solve problems is beneficial in various ways as it enables an individual to come up with a clear and concise solution to any problem or situation that may arise.

Some of the benefits that I would like to see in my own life as I begin to apply the 4-step problem-solving process are:Improved decision making skills- The 4-step problem-solving process will help me to make better decisions since it requires that I identify the problem, gather information, develop solutions, and implement the most suitable solution. This approach will help me to weigh my options carefully and make decisions based on logic rather than emotions.Improved communication skills- The process of problem-solving requires communication between team members. As I begin to apply the 4-step problem-solving process, I would like to develop better communication skills that will enable me to express my ideas clearly and succinctly.

This will be helpful not only in my personal life but also in my professional life where effective communication is a key to success.Improved critical thinking skills- The process of problem-solving requires critical thinking. By applying the 4-step problem-solving process, I would like to improve my critical thinking skills by learning how to analyze information, assess the problem, and develop solutions based on the available information.Improved productivity- The 4-step problem-solving process is a structured approach that helps to identify problems and develop solutions in a systematic manner. By applying this approach in my life, I would like to improve my productivity since I will be able to solve problems more efficiently and effectively.

for such more questions on  system

https://brainly.com/question/28021242

#SPJ8

Answer:

a) dB / dA = 2 ,

b) Network B     Network A

        2                         1

        4                         2

        6                         3

Explanation:

a) The expression for grating diffraction is

         d sin θ = m λ

where d the distance between two slits, λ the wavelength and m an integer that represents the diffraction range

In this exercise we are told that the two spectra are in the same position, let's write the expression for each network

Network A

m = 1

         sin θ = 1 λ / dA

Network B

m = 2

        sin θ = 2 λ / dB

they ask us for the relationship between the distances, we match the equations

            λ / dA) = 2 λ / dB

            dB / dA = 2

b) let's write the equation of the networks

         sin θ = m_A  λ / dA

         sin θ = m_B  λ / dB

we equalize

           m_A  λ/ dA = m_B  λ / dB

we use that

          dB / dA = 2

           m_A 2 = m_B

therefore the overlapping orders are

Network B     Network A

   2                         1

   4                         2

    6                       3

Answer:

Minimum area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum area of rectangle

Computation :

Minimum area of rectangle = Length of rectangle x Width of rectangle

Minimum area of rectangle = 6 x 4

Minimum area of rectangle = 24 centimeter²

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

For more such questions on decay rate, click on:

https://brainly.com/question/27542728

#SPJ8

The linear speed will be the insect be 0.5759 meter/second carried collision with the near stationary photograph.

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

Given that an insect lands 0.1m from the center of the turn table.

Rotational speed of the turn table = 55 rev/min

= (55×2π/60) rad/second

= 5.759 rad/second.

Hence, the speed of the insect be = Rotational speed × length

= 5.759 rad/second × 0.1 M.

= 0.5759 meter/second.

Therefore, the speed of the insect be 0.5759 meter/second.

Learn more about speed here:

https://brainly.com/question/28224010

#SPJ1

Answer:

heat from the sun comes on the earth by radiation

The amount of damping force that allow shortest possible time is called​ critical damping of the system.

Critical damping is the threshold between overdamping and underdamping  at which the oscillator returns to the position of equilibrium quickly as possible.

Critical damping is frequently desired because such a system returns to and maintains equilibrium quickly. Furthermore, a constant force applied to a critically damped system moves the system to a new equilibrium position as quickly as possible without overshooting or oscillating around the new position.

Critical damping thus provides the rapid approach to zero amplitude for a damped oscillator.

To find more on critical damping, refer here:

https://brainly.com/question/13161950

#SPJ1

Answer:

chemical energy is the answer

Answer:

Earth

lol... ....

Answer:

Explanation:

Oh sorry I thought I knew this one I guess not but you can prob look it up on safari

The total power received by the antenna is 2.14 × 10^-14 W, and  a source at a distance of 17000 ly would need to emit a power of 1.35 × 10^-26 W to provide a signal that could be detected by the radio telescope.

A radio wave is a type of electromagnetic wave with a frequency between about 3 kHz and 300 GHz. Radio waves are used for a variety of purposes, including communication, broadcasting, and radar. They are produced by oscillating electrical charges, and can travel through air and other materials, including space.

Radio waves are characterized by their frequency and wavelength. The frequency of a radio wave is the number of cycles per second, measured in hertz (Hz). The wavelength is the distance between two adjacent peaks of the wave, and is inversely proportional to the frequency. Radio waves with lower frequencies have longer wavelengths, while those with higher frequencies have shorter wavelengths.

Radio waves can be generated by a variety of sources, including antennas, electrical circuits, and certain types of electronic devices. They are used for communication between devices, such as radios and cell phones, as well as for broadcasting television and radio signals. Radio waves are also used for remote sensing, including radar and satellite communication.

Here in the Question,

(a) The power per unit area received by the antenna is given by:

P/A = 0.95 pW / 6.1 × 10^17 m^2

The total power received by the antenna is equal to the power per unit area multiplied by the area of the antenna:

P = (P/A) × π(0.5d)^2

where d is the diameter of the antenna. Substituting d = 380 m, we get:

P = (0.95 pW / 6.1 × 10^17 m^2) × π(0.5 × 380)^2

P = 2.14 × 10^-14 W

Therefore, the total power received by the antenna is 2.14 × 10^-14 W.

(b) To find the power of a source at 17000 ly distance that could provide such a signal, we need to consider the power spreading over the surface of a sphere with a radius of 17000 ly. The power density of the signal decreases with the square of the distance from the source, so the power received by the antenna from a source at a distance of 17000 ly is:

P = P0 × (d0 / d)^2

where P0 is the power emitted by the source, d0 is the distance from the source to the antenna (i.e., 17000 ly), and d is the diameter of the antenna.

Substituting the values given, we get:

P = (0.95 pW / 6.1 × 10^17 m^2) × π(0.5 × 380)^2 × (17000 × 9.461 × 10^15 m / 380 m)^2

P = 1.35 × 10^-26 W

Therefore, a source at a distance of 17000 ly would need to emit a power of 1.35 × 10^-26 W to provide a signal that could be detected by the radio telescope.

To learn more about the Doppler effect click:

https://brainly.com/question/3795295

#SPJ2

Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

answer

the first picture is c

Explanation:

because the body is moving with constant velocity

Answer:

1. The answer is c (6m)

2. The answer is a (3m)

(021) The acceleration at 1 second is the slope of the line plotted over the first 2 seconds. This line passes through the points (0 s, 0 m/s) and (2 s, 5 m/s), so its slope and thus the (average) acceleration is

(5 m/s - 0 m/s) / (2 s - 0 s) = 5/2 m/s² = 2.5 m/s²

(022) At time 2 seconds, the velocity curve passes through the point (2 s, 5 m/s), so the velocity is 5 m/s

(023) The distance traveled by this object after 2 seconds is equal to the area under the velocity function over the first 2 seconds. This region is a triangle with base 2 s and height 5 m/s, so the area is

1/2 (2 s) (5 m/s) = 5 m

The object's initial position is 10 m, so its final position after 2 seconds is

10 m + 5 m = 15 m

(024) Similar to (021): compute the slope of the line connecting the point (2 s, 5 m/s) and (6 s, 7 m/s):

(7 m/s - 5 m/s) / (6 s - 2 s) = (2 m/s) / (4 s) = 1/2 m/s² = 0.5 m/s²

(025) We know the distance traveled in the first 2 seconds. Over the next 4 seconds, the object travels an additional distance equal to the area of a trapezoid with "height" 6 s - 2 s = 4 s, and "bases" 5 m/s and 7 m/s. The area of this trapezoid is

1/2 (4 s) (5 m/s + 7 m/s) = 24 m

The net distance traveled after 6 s is then

5 m + 24 m = 29 m

so the object's position after 6 s is

10 m + 29 m = 39 m

(026) Again, compute the slope of a line, this time through the point (6 s, 0 m/s) and (9 s, -1 m/s) :

(-1 m/s - 0 m/s) / (9 s - 6 s) = (-1 m/s) / (3 s) ≈ -0.333 m/s²

(027) If the velocity at 6 s is 0 m/s, and after each second the velocity decreases by 0.333 m/s, then after 2 more seconds the velocity would be

0 m/s + (2 s) (-0.333 m/s²) = -0.666 m/s

(028) The distance traveled between the 6th second and 8th second corresponds to the area of another trapezoid, with "height" 8 s - 6 s = 2s and "bases" 0 m/s and 0.666 m/s - 0 m/s = 0.666 m/s. So the distance traveled in this interval is

1/2 (2 s) (0 m/s + 0.666 m/s) = 0.666 m

However, the velocity is negative for this duration, so the object has turned around and is moving in the opposite direction. This means it covers a net distance after a total of 8 s of

5 m + 24 m - 0.666 m = 28.333 m

and so its position after 8 s is

10 m + 28.333 m = 38.333 m

Below are the required answers and explanations for each of the scenarios listed.

1. A machine that pulls all thermal energy out of a refrigerated space: This violates the second law of thermodynamics. This is because the second law of thermodynamics states that no heat engine can have an efficiency of 100 percent, and no heat transfer can occur from a colder to a warmer object without external work being done.

2. A machine that can pull 1000J of heat out of a refrigerated space and into a warmer space without external work: This violates the second law of thermodynamics. The second law of thermodynamics states that no heat transfer can occur from a colder to a warmer object without external work being done.

3. A machine that can turn 1000J of heat directly into 1000J of electricity: This does not violate any of the laws of thermodynamics.

4. A machine that can create 1000J of heat from 100J of electricity: This does not violate any of the laws of thermodynamics.

5. A machine that can pull 1000J of heat out of a refrigerated space and put 1500J of heat into a warmer space if it uses 500J of external work: This does not violate any of the laws of thermodynamics.

a) Option 2 is correct answer.

b) Option  2 is correct answer.

c) Option 4 is correct answer.

d) Option 4 is correct answer

e) Option 4 is correct answer.

For more question  thermodynamics

https://brainly.com/question/13059309

#SPJ8

Answer:

Volume of metal piece = 0.0069 m³ (Approx.)

Explanation:

Given:

Weight of metal in air = 300 N  

Weight of metal in water = 232.5 N

Find:

Volume of metal piece

Specific gravity of metal

Computation:

We know that;

Density of water = 1,000 kg/m³

Buoyant force applied on metal piece = Weight of metal in air - Weight of metal in water

Buoyant force applied on metal piece = 300 N - 232.5 N

Buoyant force applied on metal piece = 67.5 N

Buoyant force = Volume of metal x Density of water x Gravitational force

67.5 = Volume of metal x 1,000 x 9.8

Volume of metal piece = 0.0069 m³ (Approx.)

(a) The applied upward force is 84.4 N.

(b) The tension in the massless rope is 35.4 N.

The acceleration of the boxes is calculated by applying the following kinematic equation.

v² = u² + 2as

where;

6² = 0 + 2(9)a

36 = 18a

a = 36/18

a = 2 m/s²

T - W = ma

where;

T  - (3 x 9.8) = 3 x 2

T - 29.4 = 6

T = 35.4 N

The applied upward force on block A is calculated as follows;

F = T + W

where;

F = 35.4  +  (5 x 9.8)

F = 84.4 N

Learn more about upward force here: https://brainly.com/question/10119978

#SPJ1

Explanation:

\(\underline{\boxed{\large{\bf{Option \; A!! }}}} \)

Here,

We have to find the equivalent resistance of the circuit.

Here, \(\rm { R_1} \) and \(\rm { R_2} \) are connected in series, so their combined resistance will be given by,

\(\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ \)

\(\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ \)

\(\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ \)

Now, the combined resistance of \(\rm { R_1} \) and \(\rm { R_2} \) is connected in parallel combination with \(\rm { R_3} \), so their combined resistance will be given by,

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ \)

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ \)

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ \)

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ \)

Reciprocating both sides,

\(\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ \)

Now, the combined resistance of \(\rm { R_1} \), \(\rm { R_2} \) and \(\rm { R_3} \) is connected in series combination with \(\rm { R_4} \). So, equivalent resistance will be given by,

\(\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ \)

\(\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ \)

\(\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ \)

\(\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ \)

\(\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ \)

Henceforth, Option A is correct.

\(\underline{\boxed{\large{\bf{Option \; B!! }}}} \)

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

\( \longrightarrow \) V = IR

\( \longrightarrow \) 3 = I × 3.33

\( \longrightarrow \) 3 ÷ 3.33 = I

\( \longrightarrow \) 0.90 Ampere = I

Henceforth, Option B is correct.

\( \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ \)

Answer:

500 rn

Explanation:

Answer:

Explanation:

Hay muchos recursos en línea que pueden ayudarte a aprender cómo resolver este tipo de problemas. Por ejemplo, puedes consultar el documento 1 del MIT OpenCourseWare que explica los métodos de solución para problemas electrostáticos en inglés. También puedes ver el documento 2 de ResearchGate que contiene soluciones a algunos ejercicios de electrostática en español. Otra opción es visitar el sitio web youphysics.education 3 que tiene muchos problemas y soluciones de electrostática en español.